So I'm given a general equation and am expected to get the location on the unit circle, the period, and the general solution to it. The equation is (3cot^2)2x-1 = 0. I've gotten it down to (cot^2)2x = 1/3 but I have no clue what to do next. Please help!
ok so far
(cot^2)2x = 1/3
tan^2 (2x) = 3
tan (2x) = ±√3
I recall that tan 60° = √3
so 2x = 60
2x = 120° in quad II
2x = 240° in quad III
2x = 300° in quad IV
so x = 30°, 60°, 120°, or 150° for 0 ≤ x ≤ 360°
in radians: x = π/6, π/3, 2π/3, 5π/3
take it from there
To find the location on the unit circle, the period, and the general solution to the equation (3cot^2)(2x)-1 = 0, we need to follow a few steps.
Step 1: Simplify the equation
You have already simplified it to (cot^2)(2x) = 1/3.
Note that cot^2(2x) is the square of the cotangent of 2x.
Step 2: Identify the period
The period of a trigonometric function is the length of one complete cycle. In this case, the function cot(2x) has a period of π/2.
Step 3: Solve the equation
To solve the equation (cot^2)(2x) = 1/3, we need to take the square root of both sides. However, we need to be careful because cot(2x) could be positive or negative.
Taking the square root of both sides:
√(cot^2)(2x) = ±√(1/3)
|cot(2x)| = √(1/3)
Step 4: Find the solutions
Now, consider two cases:
a) cot(2x) = √(1/3)
b) cot(2x) = -√(1/3)
Case a:
cot(2x) = √(1/3)
To find the solutions, we can take the inverse cotangent (or arccot) of both sides:
2x = arccot(√(1/3))
Since the cotangent function has a period of π/2, we need to consider values of arccot(√(1/3)) within one period.
x = (arccot(√(1/3))) / 2
Case b:
cot(2x) = -√(1/3)
Similarly, we take the inverse cotangent of both sides:
2x = arccot(-√(1/3))
Again, considering the period, we have:
x = (arccot(-√(1/3))) / 2
In both cases, you can simplify the arccot expressions if possible, for example, by using trigonometric identities or a calculator.
Step 5: Final results
The general solution to the equation (3cot^2)(2x)-1 = 0 is given by the two cases:
1) x = (arccot(√(1/3))) / 2
2) x = (arccot(-√(1/3))) / 2
These solutions represent the values of x that satisfy the equation and can locate points on the unit circle.